Answer:
Theoretical yield of the reaction is 121·38 g
The excess reactant is hydrogen
The limiting reactant is nitrogen
Explanation:
By assuming that the reaction between nitrogen and hydrogen taking place in presence of catalyst because at normal conditions the reaction between them will not occur
Number of moles of nitrogen taken are 100÷28 ≈ 3.57
Number of moles of hydrogen taken are 100÷2 = 50
Actually the reaction between nitrogen and hydrogen takes place according to the following equation
<h3>N
+ 3H
→ 2NH
</h3>
So from the equation for 1 mole of nitrogen and 3 moles of hydrogen we get 2 moles of ammonia
Here in the problem we have approximately 3·57 moles of nitrogen so we require 3×3·57 moles of hydrogen
∴ Number of moles of hydrogen required is 10·71
But we have 50 moles of hydrogen
∴ Excess reagent is hydrogen and limiting reagent is nitrogen
Number of moles of ammonia produced is 2×3·57 = 7·14
Weight of ammonia is 17 g
∴ Amount of ammonia produced is 17×7·14 = 121·38 g
∴ Theoretical yield of the reaction is 121·38 g
Answer:
Explanation:The major goals of the two rovers, according to NASA, were to determine whether life as we know it could ever have arisen on Mars (focusing particularly on searching for ancient water) and characterizing the climate and geology of
(2) Arsenic. It is an element and elements are chemically the simplest units and cannot be broken down by any chemical change.
Answer:
Well, I cannot see the options but if I were you I would choose the one closest to this. Rutherford's model shows that an atom is mostly empty space, with electrons orbiting a fixed, positively charged nucleus in set, predictable paths.
Explanation:
Again I cannot see the options but here is what I would guess. Hope this helped and have a great day! :-)
the percent yield of the reaction is 100%.
The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:
% yield = actual yield / theoretical yield * 100%
% yield of a reaction in this case Rate
In this case, the molar mass of NaBr is 102.9 g / mol, as you know:
444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g
theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g
, Replaced by the definition of percent yield:
percent yield = 728.532 grams / 728.532 grams * 100%
percent yield = 100%
Finally, the percent yield of the reaction is 100%.
<h3 />
FeBr3 is iron bromide. Also known as iron bromide. Iron bromide is an ionic compound in which iron is in a +3 oxidation state.
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